\(A=1+3+3^2+3^3+...+3^{98}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{96}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{96}\right)⋮13\).
Chứng tỏ rằng A = 1 + 3 + 3^2 + 3^3 + ... + 3^97 + 3^98 chia hết cho 13